Propagation of a wavefunction on a Riemannian sigma model

Question

I have a question about Riemannian sigma model, in particular how wavefunctions propagate. Here the Riemannian sigma model refers to the one introduced in 10.4.1 and 10.4.2 of the book $\ulcorner$ K. Hori et al., Mirror Symmetry, 2003$\lrcorner$. Let me write down the data.

• We work in an Euclidean time.
• It has a spacetime manifold $M=S^1$ and a Riemannian target $(X,g)$. On a target, we have a potential $h:X\rightarrow \mathbb{R}$.
• It has a bosonic field $\phi:M\rightarrow X$ and fermionic fields $\psi,\overline{\psi}\in \Gamma(M,\phi^*TX)$ whose degrees are shifted by $-1$ and $1$, respectively.
• The Hilbert space is a de Rham complex, i.e., $\mathcal{H}=\Omega^{\bullet}(X)$.
• Supercharges are presented as $$Q=d+dh\wedge,\quad \overline{Q}=Q^{\dagger}=d^{\dagger}+\iota_{\text{grad}h},\quad F=\text{degree of the form}, \quad H=\frac{1}{2}\{Q,\overline{Q}\}.$$

As mentioned in the beginning, I want to know how wavefunctions propagate. Consider a bosonic particle moving on a real line, with a potential $V:\mathbb{R}\rightarrow\mathbb{R}$. The Hilbert space if $L^2(\mathbb{R})$ and for $t>t_0$ we have $$\Psi(x,t)=\int_{\mathbb{R}}dx_0 K(x_0,t_0;x,t)\Psi(x_0,t_0)$$ where $K(x_0,t_0;x,t)$ has two interpretations as the path integral $\int_{\phi(t_0)=x_0, \ \phi(t)=x} D\phi \ e^{-S_E[\phi]}$ (here we set $\hbar=1$ and use Euclidean action $S_E$).

I want a similar expression for Riemannian sigma model. A naive try will be $$\Psi(x,t)=\int_{X}dx_0 K(x_0,t_0;x,t)\Psi(x_0,t_0) \quad \text{for} \quad K(x_0,t_0;x,t)=\int_{\phi(t_0)=x_0 \ \phi(t)=x} D\phi D\psi D\overline{\psi} e^{-S_E[\phi,\psi,\overline{\psi}]},$$ where when integrating over $x_0$ we use the volume form on $X$ determined by the metric $g$. However, it is a nonsense because $\Psi(x,t)\in \wedge^{\bullet}T^*_x X$ whereas $\Psi(x_0,t_0)\in \wedge^{\bullet}T^*_{x_0}X$. (Recall that $\mathcal{H}=\Omega^{\bullet}(X)$ so that wavefunctions are differential forms). To compare $\Psi(x,t)$ and $\Psi(x_0,t_0)$ we will need a connection $\nabla$ on the vector bundle $\wedge^r T^*X$ (if $\Psi(-,t_0)$ is an $r$-form). Therefore our second attempt will be $$\Psi(x,t)=\int_{X}dx_0 K(x_0,t_0;x,t)\Pi_{\phi}(\Psi(x_0,t_0)),$$ where $K(x_0,t_0;x,t)$ is defined as before, and for each path $\phi:[t_0,t]\rightarrow X$ with the boundary condition $\phi(t_0)=x_0 \ \phi(t)=x$, the map $\Pi_{\phi}:\wedge^{r}T^*_{x_0}X\rightarrow \wedge^{r}T^*_x X$ is defined to be the parallel transport along $\phi$ determined by the connection $\nabla$. This sounds reasonable, but I am not sure about it. Also I don't know which connection $\nabla$ to use. Could anyone let me know the right form of propagation?

Answer 1

The configuration space for this theory has even coordinates $x$ and odd coordinates $\psi$ (it is $\Pi TX$ as a supermanifold). The Hilbert space is (the $L^2$ completion of ) the space of function in those variables, ie $\Omega^\bullet (X)$.

The evolution operator is a unitary operator $U(t_0,t):\Omega^\bullet(X) \to \Omega^\bullet(X)$. If we want to write it as an integral kernel, it will involve an integral over the whole configuration space (over $\Pi TX$, rather than just $X$). More explicitly, with $\Psi(t_0) \in C^\infty(\Pi TX)$, we have :

\begin{align} \Psi(t,x,\psi) &= \Big[U(t,t_0)\Psi(t_0)\Big](x,\psi) \\ &= \int_{\Pi TX}\text dx_0\text d\psi_0K(x_0,\psi_0,t_0;x,\psi,t)\Psi(t_0,x_0,\psi_0) \end{align}

The kernel can be defined by a path integral as : $$K(x_0,\psi_0,t_0;x_1,\psi_1,t) = \int_{\phi(t_0) = x_0,\psi(t_0) = \psi_0}^{\phi(t) = x_1,\psi(t_1) = t_1}\mathcal D\phi\mathcal D\psi\mathcal D\bar \psi e^{-S_E(\phi,\psi,\bar\psi)}$$

We can also make sense of $K(x_0,t_0;x,t)$ as a linear map $\bigwedge^\bullet T_{x_0}^* X \to \bigwedge^\bullet T_{x}^* X$. Explicitely, for $\alpha \in \bigwedge^\bullet T_{x_0}^* X = C^\infty(\Pi T_{x_0}X)$, we have : $$(K(x_0,t_0;x,t) \alpha)(\psi) = \int_{\Pi T_{x_0}X}\text d\psi_0 K(x_0,\psi_0,t_0;x,\psi,t)\alpha(\psi_0)$$